1. LIS
O (n^2):
/*
LIS(Longest Increasing Subsequence) 最长上升子序列 O (n ^ 2)
状态转移方程:dp[i] = max (dp[j]) + 1 (a[j] < a[i],1 <= j < i)
附带有print输出路径函数
*/
void LIS(void) {
int ret = 0, last = 0;
for (int i=1; i<=n; ++i) {
dp[i] = 1; fa[i] = -1;
for (int j=1; j<i; ++j) {
if (a[j] < a[i] && dp[i] < dp[j] + 1) {
dp[i] = dp[j] + 1; fa[i] = j;
}
}
if (ret < dp[i]) {
ret = dp[i]; last = i;
}
}
printf ("%d\n", ret);
print (last); puts ("");
}
void print(int x) {
if (fa[x] != -1) {
print (fa[x]); printf (" %d", a[x]);
}
else printf ("%d", a[x]);
}
O (nlogn):
/*
LIS 二分查找优化, O(nlogn)
设当前最长递增子序列为len,考虑元素a[i]; 若d[len]<a[i],则len++,并使d[len]=a[i];
否则,在d[1~len]中二分查找第一个大于等于a[i]的位置j,使d[j]=a[i]。附上打印路径代码(准确性未知)
*/
void LIS(void) {
int len = 1; d[1] = a[1]; fa[1] = -1;
for (int i=2; i<=n; ++i) {
if (d[len] < a[i]) {
d[++len] = a[i];
pos[len] = i; fa[i] = pos[len-1];
}
else {
int j = lower_bound (d+1, d+1+len, a[i]) - d;
d[j] = a[i];
pos[j] = i; fa[i] = (j == 1) ? -1 : pos[j-1];
}
}
printf ("%d\n", len);
vector<int> res; int i;
for (i=pos[len]; ~fa[i]; i=fa[i]) res.push_back (a[i]);
res.push_back (a[i]);
for (int i=res.size ()-1; i>=0; --i) printf ("%d%c", res[i], i == 0 ? ‘\n‘ : ‘ ‘);
}
原文:http://www.cnblogs.com/Running-Time/p/4778309.html